Chapter 6, Problem 8T

To determine

To find: The value of $\sec \theta$.

Answer to Problem 8T

The value of $\sec \theta$ is $-\frac{13}{12}$.

Explanation of Solution

Given:

The value of $\sin \theta$ as $\frac{5}{13}$ and the value of $\tan \theta$ as $-\frac{5}{12}$.

Formula used:

(1) The trigonometric functions of the angle $\theta$

$\sin \theta =\frac{opposite}{hypotenuse}$

$\cos \theta =\frac{adjacent}{hypotenuse}$

$\tan \theta =\frac{opposite}{adjacent}$

(2) Reciprocal identities

$\csc \theta =\frac{1}{\sin \theta }$

$\sec \theta =\frac{1}{\cos \theta }$

$\cot \theta =\frac{1}{\tan \theta }$

Construction:

Draw a right angled triangle with opposite side as $5$, hypotenuse as $13$ and an adjacent side as $-12$.

Figure 1

Calculation:

The given trigonometric functions are $\sin \theta =\frac{5}{13}$ and $\tan \theta =-\frac{5}{12}$.

All three sides are given opposite as $5$, hypotenuse as $13$ and an adjacent as $-12$.

By reciprocal identities $\sec \theta =\frac{1}{\cos \theta }$.

To find the value of $\sec \theta$, first find value of $\cos \theta$.

The cosine function is $\cos \theta =\frac{adjacent}{hypotenuse}$

Substitute $-12$ for adjacent and $13$ for hypotenuse.

$\cos \theta =-\frac{12}{13}$

For the value of $\sec \theta$ substitute the value of $\cos \theta =-\frac{12}{13}$.

$\begin{array}{c}\sec \theta =\frac{1}{\cos \theta }\\ =\frac{1}{\frac{-12}{13}}\\ \sec \theta =-\frac{13}{12}\end{array}$

Thus the value of $\sec \theta$ is $-\frac{13}{12}$.